When we first used capacitors as feedback element of an opamp the
workings of the circuit was only looked at in terms of direct current,
charging the capacitor.
With your new knowledge of capacitive reactance, you can see how when
the input signal is an alternating current the capacitor & its
reactance control the gain of the opamp.
With low frequencies, the reactance of the capacitor is high because a
large current is stored that must be overcome each cycle in order for it
to charge in the opposite polarity. Looking at the formula, you can see
why this is true:
Xc = 1 / (2 pi f C)
With f approaching zero, the division gets larger & larger, approaching infinity.
One problem with having just a capacitor control the gain is that for
low frequencies the gain can be so high as to drive the output to
saturation on both polarities for each change in polarity of the input
signal. To prevent this, a resistor is connected in parallel to the
capacitor in order to limit the gain.
How this works is when the a low frequency is applied as input, the
reactance of the capacitor will be extremely high, & since it is in
parallel with the resistor, the equivalent resistance of the parallel
combination will always be smaller than the smallest of values, so if
the reactance is much higher than the resistance, so the resistance will
dominate (when a component dominates is when a combination tends to the
particular value of that component).
As the frequency at the input increases, the reactance of the capacitor
decreases, making the parallel combination lower & lower. This has the
effect that the ratio of Rin & Rf is smaller, making the gain of the
amplifier lower & lower, given by the equation
Vout = Vin (-Rf / Rin)
With very high frequencies, Rf is dominated by the very low reactance of
the capacitor, & the gain tends towards zero, so these frequencies
are being blocked.
As you can see, the integrator circuit is also a low pass filter,
amplifying low frequency signals & attenuating high frequency signals
to the point of blocking them